Use freezing point depression formula

$\Delta T_f = iK_f m$

For dilute aqueous solutions, comparing same solvent and same concentration, we only need to compare the van’t Hoff factor $i$.

The solution with larger $i$ will have greater depression in freezing point, hence lower freezing point.

Reference solution: $0.10\,M\; C_2H_5OH$

Ethanol is a non-electrolyte, so it does not dissociate.

$i = 1$

Check each given solution

(i) $0.10\,M\; Ba_3(PO_4)_2$

It dissociates as: $Ba_3(PO_4)_2 \rightarrow 3Ba^{2+} + 2PO_4^{3-}$ Total ions formed $= 5$

$i \approx 5 > 1$

So its freezing point is lower than that of ethanol solution.

(ii) $0.10\,M\; Na_2SO_4$

Dissociation: $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$ Total ions formed $= 3$

$i \approx 3 > 1$

So its freezing point is lower.

(iii) $0.10\,M\; KCl$

Dissociation: $KCl \rightarrow K^+ + Cl^-$ Total ions formed $= 2$

$i \approx 2 > 1$

So its freezing point is lower.

(iv) $0.10\,M\; Li_3PO_4$

Dissociation: $Li_3PO_4 \rightarrow 3Li^+ + PO_4^{3-}$ Total ions formed $= 4$

$i \approx 4 > 1$

So its freezing point is lower.

Count the number of such solutions

All four solutions have $i>1$, while ethanol has $i=1$.

Therefore, all 4 solutions have lower freezing point than $0.10\,M\; C_2H_5OH$.

Final answer

$\boxed{4}$

Comparison with stored correct answer

Stored correct answer = $4$

Our derived answer matches it.